3.11.41 \(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) [1041]

Optimal. Leaf size=154 \[ \frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

8/15*tan(f*x+e)/a^2/c^2/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+1/5*tan(f*x+e)/f/(a+I*a*tan(f*x+e)
)^(5/2)/(c-I*c*tan(f*x+e))^(5/2)+4/15*tan(f*x+e)/a/c/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3604, 40, 39} \begin {gather*} \frac {8 \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

Tan[e + f*x]/(5*f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)) + (4*Tan[e + f*x])/(15*a*c*f*(a +
I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (8*Tan[e + f*x])/(15*a^2*c^2*f*Sqrt[a + I*a*Tan[e + f
*x]]*Sqrt[c - I*c*Tan[e + f*x]])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-x)*(a + b*x)^(m + 1)*((c + d*x)^(m
+ 1)/(2*a*c*(m + 1))), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /;
FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f}\\ &=\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 3.70, size = 117, normalized size = 0.76 \begin {gather*} -\frac {\sec ^2(e+f x) (\cos (3 (e+f x))+i \sin (3 (e+f x))) (150 \sin (e+f x)+25 \sin (3 (e+f x))+3 \sin (5 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{240 a^2 c^3 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

-1/240*(Sec[e + f*x]^2*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(150*Sin[e + f*x] + 25*Sin[3*(e + f*x)] + 3*Sin
[5*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*c^3*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.35, size = 105, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \left (8 \left (\tan ^{4}\left (f x +e \right )\right )+20 \left (\tan ^{2}\left (f x +e \right )\right )+15\right )}{15 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{4}}\)	\(105\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \left (8 \left (\tan ^{4}\left (f x +e \right )\right )+20 \left (\tan ^{2}\left (f x +e \right )\right )+15\right )}{15 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{4}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^3/c^3*(1+tan(f*x+e)^2)*tan(f*x+e)*(8*tan(f*x+e
)^4+20*tan(f*x+e)^2+15)/(tan(f*x+e)+I)^4/(-tan(f*x+e)+I)^4

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Maxima [A]
time = 0.63, size = 76, normalized size = 0.49 \begin {gather*} \frac {3 \, \sin \left (5 \, f x + 5 \, e\right ) + 25 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right ) + 150 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right )}{240 \, a^{\frac {5}{2}} c^{\frac {5}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

1/240*(3*sin(5*f*x + 5*e) + 25*sin(3/5*arctan2(sin(5*f*x + 5*e), cos(5*f*x + 5*e))) + 150*sin(1/5*arctan2(sin(
5*f*x + 5*e), cos(5*f*x + 5*e))))/(a^(5/2)*c^(5/2)*f)

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Fricas [A]
time = 1.08, size = 119, normalized size = 0.77 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 28 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 175 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 175 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 28 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{480 \, a^{3} c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/480*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-3*I*e^(12*I*f*x + 12*I*e) - 28*I*e
^(10*I*f*x + 10*I*e) - 175*I*e^(8*I*f*x + 8*I*e) + 175*I*e^(4*I*f*x + 4*I*e) + 28*I*e^(2*I*f*x + 2*I*e) + 3*I)
*e^(-5*I*f*x - 5*I*e)/(a^3*c^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**(5/2)/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*(-I*c*(tan(e + f*x) + I))**(5/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(5/2)*(-I*c*tan(f*x + e) + c)^(5/2)), x)

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Mupad [B]
time = 6.13, size = 163, normalized size = 1.06 \begin {gather*} \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,125{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,22{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )-150{}\mathrm {i}\right )}{480\,a^3\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(2*e + 2*f*x)*125i + cos(
4*e + 4*f*x)*22i + cos(6*e + 6*f*x)*3i + 175*sin(2*e + 2*f*x) + 28*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x) - 150
i))/(480*a^3*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))

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